exam01_soln_cover - ME 563 – Fall 2011 Exam No. 1 Grade...

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Unformatted text preview: ME 563 – Fall 2011 Exam No. 1 Grade distribution and solution Exam scores % 93.3 93.3 93.3 92.2 92.2 91.1 90.0 90.0 88.9 87.8 86.7 86.7 83.3 82.2 81.1 81.1 81.1 81.1 80.0 80.0 78.9 77.8 72.2 70.0 67.8 67.8 62.2 61.1 61.1 58.9 57.8 56.7 56.7 56.7 56.7 25.6 ! ! Problem No. 1 (10 points) Let [A] = [aij ] represent the flexibility matrix for the three-DOF system below corresponding to generalized coordinates q1 = x, q2 = y and q3 = θ. Determine the following components of [A]: a22 , a12 , a31 and a23 . SOLUTION Apply unit force at 1: a11 = 1/k . Also, A moves same distance as O and angle of bar remains zero. a21 = a11 = 1/k = a12 and a31 = 0. Apply unit force at 2: a22 = 1/k + 1/2k = 3/2k . Also, angle of bar remains zero. Therefore a32 = 0 = a23 . Therefore, Problem No. 2 (10 points) The motion of a six-DOF system is described by the translational coordinates xj (j = 1, 2, ..., 6) as shown below. The original configuration of the system has mass and stiffness matrices of [M ] and [K ], respectively. However, we do not know any other details on the system The system is changed by: adding a spring of stiffness k between points 2 and 5, adding a spring of stiffness k between points 3 and 6, and the addition of a point mass m at point 5, all as shown in the bottom figure. If the mass and stiffness of the modified system are written as [M ]new = [M ] + [∆M ] and [K ]new = [K ] + [∆K ], determine the non-zero elements of the matrices [∆M ] and [∆K ]. Fill in these non-zero elements in the matrices provided on the next page. Provide details of your derivation in order to receive full credit for this problem. SOLUTION 1 1 1 1 ∆T = m (x2 + x5 )2 = mx2 + (2m)x2 x5 + mx2 ˙ ˙ ˙2 ˙˙ ˙5 2 2 2 2 1 1 1 = M22 x2 + (M25 + M52 )x2 x5 + M55 x2 ˙2 ˙˙ ˙5 2 2 2 Therefore, ∆M22 = ∆M25 = ∆M52 = ∆M55 = m. 1 1 1 1 1 1 ∆U = kx2 + k (x6 − x3 )2 = kx2 + kx2 − (2k )x3 x6 + kx2 5 5 6 2 2 2 2 2 23 Therefore, using∆Kij = ∂ 2 ∆U/∂ xi ∂ xj gives ∆K33 = ∆K55 = ∆K66 = k , and ∆K36 = ∆K63 = −k . Problem No. 3 (10 points) A force F acts to the right on block B of the system shown below, where x1 is the displacement of block A and x2 is the displacement of the center of the disk O. Determine the generalized forces Qx1 and Qx2 corresponding to F . SOLUTION Let θ be the rotation of disk (positive in CW sense). Therefore, θ = (x2 − x1 )/R. Using this, ￿B = (x2 + Rθ) ˆ = (2x2 − x1 ) ˆ r i i ￿r dW = F · d￿B = F ˆ · (2dx2 − dx1 ) ˆ = −F dx1 + 2F dx2 i i Therefore, Qx1 = −F and Qx2 = 2F . Problem No. 4 (10 points) A three-DOF system, whose EOMs were derived using Lagrange’s equations, has a mass matrix of: 100 [M ] = 0 1 0 002 The first two modal vectors are known: 1 1 ￿ ￿ X (1) = 3 X (2) = 3 1 −5 (3) ￿ Determine the third modal vector X (3) . Scale your answer such that X1 . SOLUTION ￿ Let X (3) 1 = a . Using orthogonality of modal vectors through [M ]: b ￿ ￿ 0 = X (1)T [M ]X (3) T 1 100 1 = 3 0 1 0 a = 1 + 3 a + 2b 1 002 b T 1 100 1 ￿ ￿ 0 = X (2)T [M ]X (3) = 3 0 1 0 a = 1 + 3a − 10b −5 002 b 1 ￿ Solving the above: a = −1/3 and b = 0. Therefore, X (3) = −1/3 0 Problem No. 5 (10 points) An undamped three-DOF system has mass and stiffness matrices of: 200 2 −1 0 2 −2 [M ] = 0 1 0 [K ] = −1 001 0 −2 1 respectively. Is the free response of the system purely oscillatory? Explain. SOLUTION K11 = 2 > 0 ￿ ￿ ￿ 2 −1 ￿ ￿ ￿=4−1=3>0 ￿ −1 2￿ ￿ ￿ ￿ 2 −1 ￿ ￿ ￿ ￿ 0￿ ￿ ￿ ￿ 2 −2 ￿ ￿ −1 −2 ￿ ￿ −1 ￿ − (−1) ￿ ￿ = −4 − 1 = −5 < 0 2 −2 ￿ = 2 ￿ ￿ ￿ ￿ −2 ￿0 1￿ 1￿ ￿ 0 −2 ￿ 1 From this we see that [K ] is not positive definite. Therefore, some of the eigenvalues will be real, and as a result, the system will some non-oscillatory modal components. Problem No. 6 (10 points) The model for an undamped three-DOF system has the following set of EOMs: ¨ [M ]￿ + [K ]￿ = ￿ x x0 This system has natural frequencies of: ω1 = 1rad/sec, ω2 = 3rad/sec and ω3 = 5rad/sec. Damping is now added to the model with the following damping matrix: [C ] = 0.06[M ] + 0.04[K ] (kg/sec) Determine the three modal damping ratios for the system. SOLUTION ￿ ￿ ￿ ￿ 1α 1 0.06 0.03 ζj = + βωj = + 0.04ωj = + 0.02ωj 2 ωj 2 ωj ωj Therefore, ζ1 = 0.03 + 0.02(1) = 0.05 1 ζ2 = 0.03 + 0.02(3) = 0.07 3 ζ3 = 0.03 + 0.02(5) = 0.106 5 Problem No. 7 (10 points) An undamped two-DOF system has a mass matrix of: ￿ ￿ 20 [M ] = 01 modal vectors of: ￿￿ 1 (1) ￿ X= 2 ￿ (2) X = ￿ 1 −1 ￿ and natural frequencies of ω1 = 0 and ω2 = √ 3rad/sec. Determine the free response of the system corresponding to initial conditions of x1 (0) = 2, x2 (0) = −2, x1 (0) = 1 and x2 (0) = 2. ˙ ˙ SOLUTION ￿ ￿ ￿ (t) = (c1 + s1 t)X (1) + (c2 cosω2 t + s2 sinω2 t)X (2) x ˙ ￿ ￿ Note that the ICs are such that ￿ (0) = 2X (2) (⇒ c1 = 0), and ￿ (0) = X (1) (⇒ s2 = 0), and: x x c2 = ￿ ￿ ￿ X (2)T [M ]￿ (0) x X (2)T [M ]X (2) =2 =2 ￿ ￿ ￿ ￿ X (2)T [M ]X (2) X (2)T [M ]X (2) s1 = ˙ ￿ ￿ ￿ X (1)T [M ]￿ (0) x X (1)T [M ]X (1) = =1 ￿ ￿ ￿ ￿ X (1)T [M ]X (1) X (1)T [M ]X (1) Therefore, ￿ ( t) = x ￿ 1 2 ￿ t+2 ￿ 1 −1 ￿ √ cos 3t Problem No. 8 (10 points) The characteristic equation for a rod problem has been found to be: 2 cotβ L = −1 βL ￿ ρ where β = E ω . • Sketch the components of the characteristic function in the plot provided below. • Place upper and lower bounds on all natural frequencies for the system. SOLUTION 2j − 1 π 2 ￿ E < ωj < j π ρL2 ￿ E ρL2 Problem No. 9 (10 points) An undamped two-DOF system with harmonic excitation has EOMs of: ￿ ￿￿ ￿￿ ￿￿ ￿￿￿ 21 x1 ¨ 4 −1 x1 1 + = sinΩt 12 x2 ¨ −1 1 x2 1 Determine all frequencies Ω corresponding to anti-resonances in the particular solution ￿ P (t). I x SOLUTION [H ] = ￿ −2Ω2 + 4 −Ω2 − 1 −Ω2 − 1 −2Ω2 + 1 ￿ −1 1 = ∆ ￿ −2Ω2 + 1 Ω2 + 1 2+1 Ω −2Ω2 + 4 where ∆ = det[H ]. Therefore, ￿ ￿￿ ￿ ￿ ￿ 2 1 Ω2 + 1 1 − Ω2 + 2 ￿ = 1 −2Ω + 1 X = Ω2 + 1 −2Ω2 + 4 1 ∆ ∆ − Ω2 + 5 From this we have: X1 = − Ω2 + 2 =0 ∆ when Ω= √ 2 X2 = − Ω2 + 5 =0 ∆ when Ω= √ 5 ￿ ...
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This document was uploaded on 12/23/2011.

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